Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
Answer: 61%
The reaction equation should be
CaF2 + H2SO4 → 2HF + CaSO4
For every 1 molecule CaF2 used, there will be 2 molecules of HF formed. The molecular mass of CaF2 is 78/mol while the molecular mass of HF is 20g/mol. If the yield is 100%, the amount of HF formed by 112g CaF2 would be: 112g/(78g/mol) * 2 * (20g/mol)=57.43g
The percentage yield of the reaction would be: 35g/57.43g= 60.94%
Answer:
The answer to your question is:
Explanation:
1 mol = 6.022 x 10 ²³ atoms
755 moles = x
x = 755 x 6.022 x 10 ²³ / 1 = 4.54 x 10 ²⁶ atoms
Uhhhhhhhhhhhhhh wut, can you rewrite it?
Answer: The volume of the gas when the temperature of the gas is raised to
is 363 ml
Explanation:
To calculate the final volume of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
![\frac{V_1}{T_1}=\frac{V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7BT_1%7D%3D%5Cfrac%7BV_2%7D%7BT_2%7D)
where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:
![V_1=281ml\\T_1=35.0^oC=(35.0+273)K=308K\\V_2=?\\T_2=125^0C=(125+273)K=398K](https://tex.z-dn.net/?f=V_1%3D281ml%5C%5CT_1%3D35.0%5EoC%3D%2835.0%2B273%29K%3D308K%5C%5CV_2%3D%3F%5C%5CT_2%3D125%5E0C%3D%28125%2B273%29K%3D398K)
Putting values in above equation, we get:
![\frac{281}{308}=\frac{V_2}{398}\\\\V_2=363ml](https://tex.z-dn.net/?f=%5Cfrac%7B281%7D%7B308%7D%3D%5Cfrac%7BV_2%7D%7B398%7D%5C%5C%5C%5CV_2%3D363ml)
Thus the volume of the gas when the temperature of the gas is raised to
if the pressure remains constant is 363 ml